Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - Cumulative Review - Page 659: 46


$\approx0.4$ seconds

Work Step by Step

The ball lands on the opposite side when $h=0$. So using the given formula with $h_0=9$ and $v_0=-16$ we get: $0=-16t^2-16t+9$ Then using the quadratic formula: $t=\frac{16\pm\sqrt{(-16)^2-4(-16)9}}{2(-16)}=\frac{-2\pm\sqrt{13}}{4}$ We consider only the positive answer: $t=\frac{2+\sqrt{13}}{4}\approx0.4$
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